RD Sharma Class 10 Solutions Chapter 14 Co-ordinate Geometry Exercise 14.4 (Updated for 2021-22)

RD Sharma Class 10 Solutions Chapter 14 Exercise 14.4: The main topic of this exercise is section formula applications. The RD Sharma Class 10 Solutions can be used to look out solutions to problems in any chapter of the book. Students can also use the links below to access the RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry Exercise 14.4 PDF.

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RD Sharma Class 10 Solutions Chapter 14 Exercise 14.4

Access answers to RD Sharma Solutions Class 10 Maths Chapter 14 Exercise 14.4- Important Question with Answers

Question 1.
Find the centroid of the triangle whose vertices are :
(i) (1, 4), (-1, -1), (3, -2)
(ii) (-2, 3), (2, -1), (4, 0)
Solution:

Question 2.
Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the Co-ordinates of the third vertex.
Solution:
Centroid of a triangle is O(0, 0) ….(i)
Co-ordinates of two vertices of a ∆ABC are A (1, 2) and B (3, 5)
Let the third vertex be (x, y)

Question 3.
Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.
Solution:
Let two vertices of a ∆ABC be A (-3, 1) and B (0, -2) and third vertex C be (x, y)
Centroid of the ∆ABC is O (0, 0)

Question 4.
A (3, 2) and B (-2, 1) are two vertices of a triangle ABC whose centroid G has the coordinates (53 , −13) . Find the coordinates of the third vertex C of the triangle. [CBSE 2004]
Solution:
A (3, 2) and B (-2, 1) are the two vertices of ∆ABC whose centroid is G (53 , −13)
Let third vertex C be (x, y)

Question 5.
If (-2, 3), (4, -3) and (4, 5) are the mid-points of the sides of a triangle, find the co-ordinates of its centroid.
Solution:
In ∆ABC, D, E and F are the mid-points of the sides BC, CA and AB respectively.
The co-ordinates of D are (-2, 3), of E are (4,-3) and of F are (4, 5)
Let the co-ordinates of A, B and C be (x₁, y₁), (x₂, y₂), (x₃, y₃) respectively

Question 6.
Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.
Solution:
In ∆ABC,
D and E are the mid points of the sides AB and AC respectively

Question 7.
Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another.
Solution:
Let A (x₁, y₁), B (x₂, y₂), C (x₃, y₃) and D (x₄, y₄) be the vertices of quadrilateral ABCD
E and F are the mid points of side BC and AD respectively and EF is joined G and H are the mid points of diagonal AC and BD.
GH are joined

Question 8.
If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA² + PB² + PC² = GA² + GB² + GC² + 3GP².
Solution:
In AABC, G is the centroid of it Let P (h, x) is any point in the plane
Let co-ordinates of A are (x₁, y₁) of B are (x₂, y₂) and of C are (x₃, y₃)


Hence proved.

Question 9.
If G be the centroid of a triangle ABC, prove that AB² + BC² + CA² = 3 (GA² + GB² + GC²)
Solution:
Let the co-ordinates of the vertices of ∆ABC be A (x₁, y₁), B (x₂, y₂), C (x₃, y₃) and let G be the centroid of the triangle



Hence proved.

Question 10.
In the figure, a right triangle BOA is given. C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A and B.
Solution:

In right ∆OAB, co-ordinates of O are (0, 0) of A are (2a, 0) and of B are (0, 2b)
C is the mid-point of AB
Co-ordinates of C will be

We see that CO = CA = CB
Hence C is equidistant from the vertices O, A and B.
Hence proved.

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